Given "f(x)=x^6"
Tangent line to "f(m)" of "x=4"
at "x = 4 :\n\ny = x^6;\ny = 4^6 \\implies" "(4, 4^6)"
Slope: "m(y) = y`= 6*5"
"m = 6*4^5"
"y - y_o = m(x-x_o)"
"y - 4^6 = 6*4^5[x-4]\\\\\n\n\\\\y = x[6*4^5] - 24*4^5+46\\\\y=6*4^5[x] - 4^5*(24-4)"
"y = 6*4^5[x] - 20*4^5"
So compaing with y = m*x+c:
m = "6(4)^5;" c = "-20(4)^5"
To estimate value of "3.7^6" using linear approximation:
f(x) = f(a)+f`(a)(x-a)
so "f(3.7) = f(4) + f`(4)(3.7-4) = 4^6+6*4^5*(-0.3)"
"= 4^6 - 6*(0.3)*(4)^5 = 2252.8"
So, linear approximation to estimate "3.7^6" is 2252.8
Comments
Leave a comment